How do I derive a general formula for $I(n,m)$? 
To prove the first one, here's what I did:
$\int x^nln(x)^mdx=$ 
IBP: $ln(x)^m=u \Rightarrow \frac{mln(x)^{m-1}}{x}dx=du$
$x^ndx=dv \Rightarrow v=\frac{x^{n+1}}{n+1}$ $\Rightarrow$
$$ln(x)^m\frac{x^{n+1}}{n+1}-\int\frac{x^{n+1}}{n+1}\frac{mln(x)^{m-1}}{x}dx=ln(x)^m\frac{x^{n+1}}{n+1}-\frac{m}{n+1}\int x^nln(x)^{m-1}dx=ln(x)^m\frac{x^{n+1}}{n+1}-\frac{m}{n+1}I(n,m-1)$$
To show the second:
$I(n,1)=\frac{x^{n+1}}{n+1}ln(x)-\frac{1}{n+1}I(n,0)$
$I(n,0)=\int x^nln(x)^{0}= \int x^n= \frac{x^{n+1}}{n+1} \Rightarrow I(n,1)=\frac{x^{n+1}}{n+1}ln(x)-\frac{1}{n+1}\Biggl( \frac{x^{n+1}}{n+1}\Biggr)$
Now, for the third:
$I(0,1)=xln(x)-I(0,0)=x\Bigl(ln(x)-1\Bigr)$ since $I(0,0)=\int x^0ln(x)^0= \int 1= x$
$I(0,2)=xln(x)^2-I(0,1)=xln(x)^2-x(ln(x)-1)=x\Biggl(ln(x)^2-ln(x)+1\Biggr)$
$I(0,3)=xln(x)^3-x(ln(x)^2-ln(x)+1)= x\Biggl(ln(x)^3-ln(x)^2+ln(x)-1\Biggr)$
$\Rightarrow I(0,m)=x(ln(x)^m+(-1)^mln(x)^{m-1}+...+(-1)^m)$
Would this be correct? 
Also how do I derive a general formula for $I(n,m)$?
 A: Knowing $I(n,0)$ and how to express $I(n,m)$ in terms of $I(n,m-1)$ means you can express $I(n,m)$ in terms of $I(n,0)$ by unfolding the recursion. You can write this recursion as
$$I(n,m)=f_{n,m}+c_{n,m-1} I(n,m-1)$$
where according to your work $f_{n,m}(x)=\ln(x)^m \frac{x^{n+1}}{n+1}$ and $c_{n,m-1}=-\frac{m}{n+1}$.
Unfolding one more time just to make the pattern more apparent:
$$I(n,m)=f_{n,m}+c_{n,m-1} f_{n,m-1} + c_{n,m-1} c_{n,m-2} I(n,m-2).$$ 
So you see that $f_{n,m-k}$ will have $k$ of the $c$'s multiplying with it when you are done unfolding, and then you will sum up all those terms. So $I(n,m)=\sum_{k=0}^m \left ( \prod_{j=1}^k c_{n,m-j} \right ) f_{n,m-k}$ for $m \geq 1$. You can simplify the product inside the sum down to be in terms of factorials and exponents.
A: You start iterating the recurrence like this:
\begin{align*}
& I(n,m) = \frac{{x^{n + 1} }}{{n + 1}}\log ^m x - \frac{m}{{n + 1}}I(n,m - 1)
\\ &
 = \frac{{x^{n + 1} }}{{n + 1}}\log ^m x - \frac{m}{{n + 1}}\left[ {\frac{{x^{n + 1} }}{{n + 1}}\log ^{m - 1} x - \frac{{m - 1}}{{n + 1}}I(n,m - 2)} \right]
\\ &
 = \frac{{x^{n + 1} }}{{n + 1}}\log ^m x - \frac{{x^{n + 1} }}{{(n + 1)^2 }}m\log ^{m - 1} x  + \frac{1}{{(n + 1)^2 }}m(m - 1)I(n,m - 2)
\\ & = \frac{{x^{n + 1} }}{{n + 1}}\log ^m x - \frac{{x^{n + 1} }}{{(n + 1)^2 }}m\log ^{m - 1} x \\ & \quad \, + \frac{1}{{(n + 1)^2 }}m(m - 1)\left[ {\frac{{x^{n + 1} }}{{n + 1}}\log ^{m - 2} x - \frac{{m - 2}}{{n + 1}}I(n,m - 3)} \right]
\\ & = \frac{{x^{n + 1} }}{{n + 1}}\log ^m x - \frac{{x^{n + 1} }}{{(n + 1)^2 }}m\log ^{m - 1} x + \frac{{x^{n + 1} }}{{(n + 1)^3 }}m(m - 1)\log ^{m - 2} x \\ &\quad\,- \frac{1}{{(n + 1)^2 }}m(m - 1)\frac{{m - 2}}{{n + 1}}I(n,m - 3),
\end{align*}
etc. From this, you can conjecture that
$$
I(n,m) = \frac{{x^{n + 1} }}{{n + 1}}\sum\limits_{k = 0}^m {\frac{{( - 1)^k }}{{(n + 1)^k }}\frac{{m!}}{{(m - k)!}}\log ^{m - k} x} ,
$$
and verify it by using the recurrence and induction.
